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# Degree 1 Nédélec (second kind) on a tetrahedron

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In this example:
• $$R$$ is the reference tetrahedron. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle z\\\displaystyle 0\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle z\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 0\\\displaystyle z\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{11}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{t}}_{0}$$ is the tangent to edge 0.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle 2 z\\\displaystyle 4 y\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{t}}_{0}$$ is the tangent to edge 0.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle 0\\\displaystyle - 4 z\\\displaystyle - 2 y\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{1}$$
where $$e_{1}$$ is the 1st edge;
and $$\hat{\boldsymbol{t}}_{1}$$ is the tangent to edge 1.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle 2 z\\\displaystyle 0\\\displaystyle 4 x\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{1}$$
where $$e_{1}$$ is the 1st edge;
and $$\hat{\boldsymbol{t}}_{1}$$ is the tangent to edge 1.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle - 4 z\\\displaystyle 0\\\displaystyle - 2 x\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{4}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
and $$\hat{\boldsymbol{t}}_{2}$$ is the tangent to edge 2.

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle 2 y\\\displaystyle 4 x\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{5}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
and $$\hat{\boldsymbol{t}}_{2}$$ is the tangent to edge 2.

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle - 4 y\\\displaystyle - 2 x\\\displaystyle 0\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{6}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{3}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{3}$$
where $$e_{3}$$ is the 3th edge;
and $$\hat{\boldsymbol{t}}_{3}$$ is the tangent to edge 3.

$$\displaystyle \boldsymbol{\phi}_{6} = \left(\begin{array}{c}\displaystyle - 2 z\\\displaystyle - 2 z\\\displaystyle - 4 x - 4 y - 6 z + 4\end{array}\right)$$

This DOF is associated with edge 3 of the reference element.
$$\displaystyle l_{7}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{3}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{3}$$
where $$e_{3}$$ is the 3th edge;
and $$\hat{\boldsymbol{t}}_{3}$$ is the tangent to edge 3.

$$\displaystyle \boldsymbol{\phi}_{7} = \left(\begin{array}{c}\displaystyle 4 z\\\displaystyle 4 z\\\displaystyle 2 x + 2 y + 6 z - 2\end{array}\right)$$

This DOF is associated with edge 3 of the reference element.
$$\displaystyle l_{8}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{4}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{4}$$
where $$e_{4}$$ is the 4th edge;
and $$\hat{\boldsymbol{t}}_{4}$$ is the tangent to edge 4.

$$\displaystyle \boldsymbol{\phi}_{8} = \left(\begin{array}{c}\displaystyle - 2 y\\\displaystyle - 4 x - 6 y - 4 z + 4\\\displaystyle - 2 y\end{array}\right)$$

This DOF is associated with edge 4 of the reference element.
$$\displaystyle l_{9}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{4}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{4}$$
where $$e_{4}$$ is the 4th edge;
and $$\hat{\boldsymbol{t}}_{4}$$ is the tangent to edge 4.

$$\displaystyle \boldsymbol{\phi}_{9} = \left(\begin{array}{c}\displaystyle 4 y\\\displaystyle 2 x + 6 y + 2 z - 2\\\displaystyle 4 y\end{array}\right)$$

This DOF is associated with edge 4 of the reference element.
$$\displaystyle l_{10}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{5}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{5}$$
where $$e_{5}$$ is the 5th edge;
and $$\hat{\boldsymbol{t}}_{5}$$ is the tangent to edge 5.

$$\displaystyle \boldsymbol{\phi}_{10} = \left(\begin{array}{c}\displaystyle - 6 x - 4 y - 4 z + 4\\\displaystyle - 2 x\\\displaystyle - 2 x\end{array}\right)$$

This DOF is associated with edge 5 of the reference element.
$$\displaystyle l_{11}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{5}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{5}$$
where $$e_{5}$$ is the 5th edge;
and $$\hat{\boldsymbol{t}}_{5}$$ is the tangent to edge 5.

$$\displaystyle \boldsymbol{\phi}_{11} = \left(\begin{array}{c}\displaystyle 6 x + 2 y + 2 z - 2\\\displaystyle 4 x\\\displaystyle 4 x\end{array}\right)$$

This DOF is associated with edge 5 of the reference element.